Question

Review Problem 6.057 Incorrect. Did you use the correct temperatures for Initial and final states? Check the signs of your answers. A 6.57 g mass of a metal was heated to 105.70°C and then plunged into 100.0 g of water at 21.36°C. The temperature of the resulting mixture became 24.91 a) How many joules did the water absorb? joules b) How many Joules did the metal lose? joules c) What is the heat capacity of the metal sample? d) What is the specific heat of the metal? -...5.01.01 Show Work is REQUIRED for this question: Open Show Work

Answer 1

Amount of heat absorbed or evolved (q) = mc∆T = mc(T_{​​​​​​2} - T_{​​​​​​1})

Where m is the mass of the sample

c is the specific heat of the sample

T_{​​​​​​2} is the final temperature

T_{​​​​​1} is the initial temperature

Given that mass of the metal = 6.57 g

Mass of the water = 100 g

Initial temperature of the metal = 105.70°C

Final temperature of the metal = 24.91°C

Initial temperature of the water = 21.36°C

Final temperature of the water = 24.91°C

Amount of heat absorbed by the water

= 100 g * (4.184 J \°C-g )* (24.91°C -21.36°C)

= 1485.32 J

Since Amount of heat lost by the metal = Amount of heat absorbed by the water,

Amount of heat lost the by the metal = -1485.32 J

Specic heat of the metal :

q = mc∆T

-1485.32 J = 6.57 g * c * (24.91 °C - 105.7 °C)

Specific heat of the metal = -1485.32 J/ (6.57 g * -80.8 °C)

= 2.798 J/ °C-g

Heat capacity of the metal

= Specific heat capacity* mass of the metal

= (2.798 J/ °C -g )*6.57 g

= 18.383 J/°C